Question

Part of a usability study of programmable thermostats, the time on task for a series of task performed for each user was recorded.

A few background items:

the participants/users are a homogeneous group.

the participants were randomly assigned to each group.

the to items 1 and 2 above, assume the two groups of participants have equal variances.

the have no information to indicate that one time on task for one group will be faster than the other.

Your job will be to perform a “t” test on these data and draw whatever conclusions you believe you can get from the data. If you need a refresher of the “t” test, read the “t-test description.pdf” document. If you need more information, check your statistics book, or use the Internet to find web sites such as http://www.graphpad.com/quickcalcs/ttest1.cfm. (Excel has a “t” test function although it may not be currently installed in your version; you would then add it in.)

Question1 – What is the null hypothesis in this evaluation? (Discussed in class, but easy found on the Internet)

Question 2 – Which “t” test should be used – paired, unpaired/equal variance, unpaired/unequal variance?

Question 3 – Should a one-tail, or two-tailed test be used, and why?

Question 4 – Is the t value significant at the 0.05 level, and why?

Question 6 – Based on (1) the above analysis, and (2) the number of ballots completed (see spreadsheet), what, if anything can you conclude?

Group 1	Group 1
72	130
31	814
68	259
74	106
62	127
28	150
87	139
71	103
93	151
80	108
59	141
65	80
59	50
35	64
69	91
54	138
45	103
41	77
68	95
40	95
38	161
82	140

Answer 1

1)  the null hypothesis in this evaluation is

or,

2) We should use Unpaired equal variance t test.

3) Here we use two-tailed test. because, We have to check which group will be faster.

4)

Group 1	Group2	Group(1-2)
72	130	-58
31	814	-783
68	259	-191
74	106	-32
62	127	-65
28	150	-122
87	139	-52
71	103	-32
93	151	-58
80	108	-28
59	141	-82
65	80	-15
59	50	9
35	64	-29
69	91	-22
54	138	-84
45	103	-58
41	77	-36
68	95	-27
40	95	-55
38	161	-123
82	140	-58
60	151	-91

t test statistic is given by,

where, s2 =pooled variance = 12069.64

t = -2.746

t critical value = 2.018 < |t|

And p-value = 0.009 < 0.05

Hence null hypothesis is rejected.

We can perform this test in excel.

	Group 1	Group 2
Mean	60	151
Variance	352.33	23786.95
Observations	22	22
Pooled Variance	12069.64
Hypothesized Mean Difference	0
df	42
t Stat	-2.746
P(T<=t) two-tail	0.009
t Critical two-tail	2.018

6) t value is significant at level 0.05 , Hence, Group 2 is faster than group 1 one at a time.

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